The solubility of sodium trioxonitrate(V), \text{NaNO}_3, in water is 4.5\text{ mol dm}^{-… — JAMB UTME

The solubility of sodium trioxonitrate(V), $\text{NaNO}_3$ , in water is $4.5\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $2.0\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . If the molar mass of $\text{NaNO}_3$ is $85\text{ g mol}^{-1}$ , what mass of $\text{NaNO}_3$ in grams will crystallize out when $400\text{ cm}^3$ of a saturated solution of the salt at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ?

Answer: 85 g

Answer

The mass of

\text{NaNO}_3

that will crystallize out is

85\text{ g}

The difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

4.5 - 2.0 = 2.5\text{ mol dm}^{-3}

. In

400\text{ cm}^3

(

0.4\text{ dm}^3

) of solution, the amount of salt precipitated is

2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

. Multiplying by the molar mass (

85\text{ g mol}^{-1}

) gives

85\text{ g}

Step-by-Step Solution

Calculate the difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

\Delta S = 4.5\text{ mol dm}^{-3} - 2.0\text{ mol dm}^{-3} = 2.5\text{ mol dm}^{-3}

Cooling causes the excess solute to precipitate out based on the difference in saturation concentration.

Convert solution volume to

\text{dm}^3

V = \frac{400\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.4\text{ dm}^3

Solubility is given per

\text{dm}^3

, so volume must be in

\text{dm}^3

Calculate the moles of solute precipitated

n = 2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

Multiplying the concentration difference by the solution volume yields total precipitated moles.

Convert moles to mass in grams

m = 1.0\text{ mol} \times 85\text{ g mol}^{-1} = 85\text{ g}

Mass is found by multiplying moles by molar mass.

Key Concept

Crystallization and solubility change with temperature

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